Understanding if $\sin x = t$ then $\cos x dx = dt$ Hi I need help in understanding this:
if $\sin x = t$ then $\cos x dx = dt$.
My math book mostly uses the Lagrange's notation (prime) and I think I may not fully have grasped the $\frac{dx}{dy}$ way of writing and calculating the derivative of a function, since I am having a hard time to truely understand the statement above. 
If anyone can give some explanation on how this logic works, I would be more than happy :)
Thank you!
 A: $$t=\sin\ x \implies t+dt=\sin(x+ dx)=\sin x\cos dx+\cos x \sin dx$$
$$dx \approx 0\implies \sin dx \approx dx \ \text{ and } \ \ \cos dx \approx 1$$ Here the property ($ \sin x \approx x \text{ when } x\approx 0$) was applied, hence $\sin dx \approx dx$
$$t+dt=\sin x\cos dx+\cos x \sin dx = \sin x + \cos x \cdot dx$$ $$\text{$\sin x =t$ so $t+dt= t+ \cos x \cdot dx\implies dt = \cos x\cdot dx$}$$
This is the logic behind it but to not disturb yourself in the future, multiply both sides by $dx$, $$\cfrac{df}{dx} = g(x)\implies df=g(x)\cdot dx$$
A: Let $y=f(x)$ be a real function. If $f(x)$ is differentiable at  $x_0$, then the the expression $$dy=f'(x_0)dx$$ is called the differential of $f$ at $x_0$. 
Or using the Leibniz's notation for the derivatives
$$dy=\left. \frac{df}{dx}\right\vert _{x_{0}}\; dx$$
For a generic $x$, we thus have
$$dy=f'(x)dx=\frac{df}{dx} dx$$
In the picture below this equation in $dx,dy$ represents the tangent line to the graph of $f(x)$ at $x_0$ in the translated coordinates system $dx,dy$.
Both $dy$ and $dx$ are interpreted as infinitesimals. The differential $dy$ is approximately the change of $y$ when $x$ changes by an arbitrary small quantity $dx$. 

In the present case we have the function $t=f(x)=\sin x$, whose derivative is 
$$t^{\prime }=f^{\prime }(x)=\frac{df}{dx}=\cos x.$$ So, the diferential $dt$ is 
$$dt=f^{\prime }(x)dx=\frac{df}{dx}dx=\cos x\;dx.$$
Further information on the answer to the question What is, how do you use, and why do you use differentials? What are their practical uses?
A: If $t=\sin x$ then $t$ is a function of $x$ so $\frac{\mathrm{d}t}{\mathrm{d}x}=\cos x$ and hence $\mathrm{d}t=\cos x\mathrm{d}x$.
